Quote:
Originally Posted by bkahl
Please elaborate.
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Alright, so first let me lay down some qualifiers.
I'm going to assume that the amperage never exceeds either 40A or 30A on the snap breakers, and that power scales linearly to acceleration. Further, there's no mechanical loss of power due to friction ect.
In it's simplest form, a DC motor is really just a square loop of current (multiple loops) inside of a Magnetic Field (referred to as B-field from this point on). Now the equation for Force when there is a line of current and a magnetic field is

(Equation 1)
What's important is that the Force is proportional to current flowing through the wire, assuming that the length of the loop and B field stay the same. Since the current flows in opposite directions on each side of the loop, the force produced by each side sums.
Torque is equal to r x F (radius cross force) (I'm skipping directions for the vectors because I don't have the time to draw everything out) or r*F.
Now, Power in rotational motion is Prot = T*omega (Torque*Rotational Velocity)
I'm going to pull from
this webpage. You can go there to see how this next equations/graph were derived.
Power for a DC motor with respect to speed and torque end up being quadratics.
and they produce this graph
This curve is for a Green Maxon Motor, but the CIM shares a very similar power curve (as do many DC motors).
We know that Power is dependent upon both Torque and Omega. We also know that Torque is equal to rI*integral dlxB.
So, assuming that B, dl (or essentially L), r, and omega remain the same, power is affected by changes in the magnitude of current.
If we limit the CIM to 40Amps, it will produce a peak power. However, if the CIM is limited to 30Amps, it will produce a lower peak power because it has less current flowing through the loop that creates the torque.
However, our power systems aren't limited to only 40 amps, you can exceed the amperage on our breakers by quite a bit. I have no idea exactly how much a 30A vs 40A breaker will affect overall power output, but it's enough that we have noticed acceleration differences.
So barring any of the current limitations discussed above, what else puts a swerve down on power compared to other FRC drive-trains (primarily 4/6/8wd to keep it simple)? Here's the main one that comes to mind.
In a 6wd all the wheels are chained together, and one gearbox of (usually) multiple motors powers all the wheels on that side. So that when your robot is being lifted/losing normal force (under defense ect.) you don't lose the power. Since (most) swerves now days have an individual module per wheel (and a single motor), this benefit is lost. When the front is off the ground, the power being produced by those wheels is simply lost. Whereas in a 6wd, that power is still being put to the ground by the rear 1-2 wheels on each side.
Obviously, this ignores things like slippage of the wheels, drivetrain ineffiencies, and other trade-offs that make it a bit more complicated. Power also doesn't scale linearly to more acceleration iirc, so keep that in mind as well.
Edit:
Let's say each swerve module has 1 CIM, this cim produces a power P.
When all 4 wheels are on the ground, the drivetrain is producing 4P.
When 1-2 wheels are off the ground, the drivetrain is producing 2-3P.
With a 6wd, assuming each side has 2 motors, the drivebase has a power of 4P. When 2 wheels are off the ground, the base is still producing power 4P because the wheels are connected and at least one is still touching the ground. (again, this is simplified a bit)